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Let $G$ be a finite group and $S$ a finite set of prime numbers. I know that every separable $\mathbb Q$-algebra $A$ contains a maximal $\mathbb Z$-order but I wonder if the following is true.

Is there a $\mathbb Z$-order $\Lambda$ in $\mathbb Q[G]$ which contains $\mathbb Z[G]$ and satisfies the following two conditions

1) $\Lambda\otimes_\mathbb Z \mathbb Z_p$ is a maximal $\mathbb Z_p$-order in $\mathbb Q_p[G]$ if $p\in S$

2) $\Lambda\otimes_\mathbb Z \mathbb Z_p=\mathbb Z_p[G]$ if $p\notin S$?

Here $\mathbb Q_p$ (resp. $\mathbb Z_p$) is the $p$-adic completion of $\mathbb Q$ (resp. $\mathbb Z$).

A proof or reference will be gratefully accepted.

Many thanks.

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  • $\begingroup$ Yes, given an order $\Lambda$, for any choice of local orders $\Lambda_p'$ that are equal to $\Lambda_p$ at all but finitely many $p$, there exists an order $\Lambda'$ whose completion at every prime $p$ is $\Lambda_p'$. $\endgroup$
    – Aurel
    Apr 21, 2016 at 17:25
  • $\begingroup$ How would this be shown? regards $\endgroup$
    – eddie
    Apr 22, 2016 at 7:24
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    $\begingroup$ (note that I was implicitly assuming the condition $\Lambda_p'\otimes\mathbb{Q}_p = \Lambda_p\otimes\mathbb{Q}_p$) Let $A = \Lambda\otimes\mathbb{Q}$ and $X = A\otimes\widehat{\mathbb{Z}}$ (finite adèles of $A$). Then define $\Lambda' = A\cap \prod_p \Lambda_p'$, where this intersection is taken in $X$. This is a very standard technique; I will try to find a reference for you. $\endgroup$
    – Aurel
    Apr 22, 2016 at 7:44

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